Can someone check these answers for me? Has to do with radicals.
Umm.. What are your answers?
They're directly under the problems. If you give me a second I can circle them and re-upload.
Oh i see them now.
Number 1 is wrong.
I thought so, but I wasn't sure how to do it.
Ok. Simply plug in your m value. Your equation is basically asking what is the square root of (4)(m).
After that I'm confused on how to actually write the answer to that. I know what the decimal says but I don't know how to write it the other way.
decimal?
As in the number after you use the calculator to do it. It comes out to like 12.21 rounded. Because I did 4*36=144, and then 144+5, and then I did square root of 149 on the calculator and it was 12.206
Ah... Do you know what the order of operations are?
yeah, pemdas.
Remember this. Exponent is also the same as square root(its recipical) in pemdas.
Oh I didn't know that.
So I would do 4*36 first, and find the square root, then add 5?
indeed.
So let me know when you get the answer. Also number 6 is wrong. Everything else looks fine to me
So if I do that, it is 17. Then would it really be like something\[\sqrt{4}\]
like some number on the outside.
nope. There is nothing outside of 17. 17 is your answer.
Without the square root sign though, right?
Yep. When you got 12 from the square root of 144, it simplified your square into a whole number
Okay, thanks. You're very helpful :) I'm gonna try 6 and let you know.
I actually don't see how 6 is wrong.
Ok so what do you get when you multiply 51 and 5?
255
So when you simplify it what do you end up with. Leave the 2 out of the question.
15.97
Do you see where the problem is? Your answer needs to end with the square root of 51.
What is the point of the other numbers then? I don't really understand.
So question 6 is asking you to find x. So your equation is 2rad51x=10rad51. Find x. The numbers below are just showing u what could be an answer.
Right. I put both of the equations into google's calculator and got the same answer by substituting the x for 5. That's why I thought it was right.
By other numbers I meant the number on the outside of the radical sign. I know that sounds completely ridiculous but oh well.
Well it is possible that you entered in the problem wrong. And google is known to not be perfect. for the equation hint: you do not need to completely multiply an answer if it is going to be squared. ex. square root of 150= square root of 15*square root of 10.
Oh i get what you are talking about. the numbers in front of the radical?
Yes. I watched the lesson and everything but I don't understand what to do with it or what it means.
Well it just means that u multiply that number by the square root. its basically this. 2* rad51
So for the problem it wants it to be equal to it's 10* rad51. So I don't try to find the square root of 51 and then multiply that by 10?
It's 25 I think.
well think of it this way. you have rad 51 already and you need it for your answer right?
right.
You have a two u can multiply to get 10. So what number times 2 do you need to get 10?
5
and knowing that your radical can be expressed as rad51* radx, what does your x have to equal to get you your answer?
@aperfectcircle
So are you saying the number used to make 2 into 10 is the same number as x?
I'm sorry, I'm just literally stuck.
No. Im saying that the square root of x is used to make 2 into 10. So you know that you need a 5. And your equation right now is 2*rad51*radx.
well 5 is the square root of 25
go on
you are so close. All you need to do is do x= something.
I did something earlier and I don't remember what I did. Somewhere up there I said I thought it was 25. I'm trying to replicate how I did that.
Well 25 is your answer. Let me know if you get stuck on something. i know my teachings are long but I try to make sure people understand how they can do it. So let me know if u get stuck on something.
I really appreciate it. I got it somewhere up there but I really don't remember how I did it. I'm gonna stop for a minute and refresh my brain and maybe it'll come back to me. Thanks so much for your help.
No problem. Also it doesnt hurt to read what we just did to solve your equation.
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